Palindromes In Base 11

Number Theory Level 3

Let n n be a m m -digit palindromic number (where m ≥ 2 m\geq 2 is even) in base 10.

What is the units digit of n n when written out in base 11?

If you think the answer depends on the value of n n in base 10, enter -1 as your answer.

Details and Assumptions:

  • The digits allowed in base 11 are digits 0-9 and A where A ‾ 11 = 10 ‾ 10 \overline{A}_{11}=\overline{10}_{10} .

  • If you think the answer depends on the value of n n in base 10, enter -1 as your answer.

  • If you think the answer is one of the digits 0-9, enter that digit as your answer. If you think the answer is A, enter 10 as your answer.


The answer is 0.

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Prasun Biswas by Prasun Biswas , 23, India

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1 solution

Chew-Seong Cheong
Jan 26, 2016

We note that n n in base 10 10 can be represented as:

n = ∑ k = 1 m 2 a k ( 1 0 m − k + 1 0 k − 1 ) , where a k = 0 , 1 , 2 , 3...9 and a 1 ≠ 0 = ∑ k = 1 m 2 a k ( 1 0 k − 1 ) ( 1 0 m − 2 k + 1 + 1 ) = ∑ k = 1 m 2 a k ( 1 0 k − 1 ) ( 10 + 1 ) ( 1 0 m − 2 k − 1 0 m − 2 k − 1 + 1 0 m − 2 k − 2 − . . . + 1 ) = 11 ∑ k = 1 m 2 a k ( 1 0 k − 1 ) ( 1 0 m − 2 k − 1 0 m − 2 k − 1 + 1 0 m − 2 k − 2 − . . . + 1 ) \begin{aligned} n & = \sum_{k=1}^\frac{m}{2} a_k \left(10^{m-k} + 10^{k-1} \right) \text{, where } a_k = 0,1,2,3...9 \text{ and } a_1 \ne 0 \\ & = \sum_{k=1}^\frac{m}{2} a_k \left( 10^{k-1} \right) \left(10^{m-2k+1} + 1 \right) \\ & = \sum_{k=1}^\frac{m}{2} a_k \left(10^{k-1} \right) \left(\color{#3D99F6}{10 + 1} \right) \left(10^{m-2k} -10^{m-2k-1} + 10^{m-2k-2} - ... + 1 \right) \\ & = \color{#3D99F6}{11} \sum_{k=1}^\frac{m}{2} a_k \left(10^{k-1} \right) \left(10^{m-2k} -10^{m-2k-1} + 10^{m-2k-2} - ... + 1 \right) \end{aligned}

We note that n n is divisible by 11 11 , therefore n n in base 11 11 has 0 \boxed{0} as its unit digit.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

That is quite impressive! Nice solution..

Zeeshan Ali - 5 years, 4 months ago

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